The two spheres of equal mass m are able to slide along the horizontal rotating
ID: 1718197 • Letter: T
Question
The two spheres of equal mass m are able to slide along the horizontal rotating rod. If they are initially latched in position a distance r from the rotating axis with the assembly rotating freely with an angular velocity wo, determine the new angular velocity w after the spheres are released and finally assume positions at the ends of the rod at a radial distance of 3.Or. Also find the fraction n of the initial kinetic energy of the system which is lost. Neglect the small mass of the rod and shaft.Explanation / Answer
>>As, L = Iw,
where,
L = Angular Momentum
I = Moment of Inertia
w = Angular Velocity,
>> As, I = mr2 + mr2 = 2*mr2
=> L = 2*mr2*w
>> As, there is no external moment acting,
So, ANgular Momentum is conserved
>> Initially, distance = r and w = wo
>> Finally, distance = 3r and Le angular velocity = w
=> 2*mr2*wo = 2*m(3r)2*w
=> w = Final Angular Velocity = wo/9 = 0.11*wo .......ANSWER.........
>> As, Kinetic Energy, K.E. = (1/2)*I*w2
=> K.E.i = Initial Kinetic Energy = 0.5*2*mr2*wo2 = mr2*wo2 = Ei
=> K.E.f = Final Kinetic Energy = 0.5*2*m(3r)2*w2 = m(3r)2*(wo/9)2 = Ei/9
So, Kinetic Energy Lost = K.E.i - K.E.f = (8/9)K.E.i
=> n = 8/9 = 0.889 , i.e. 0.889 or 88.9% of Kinetic Energy is lost .........ANSWER........
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